$ 8^{-\frac{4}{3}}$
Solution: $= \left(\dfrac{1}{8}\right)^{\frac{4}{3}}$ $= \left(\left(\dfrac{1}{8}\right)^{\frac{1}{3}}\right)^{4}$ To simplify $\left(\dfrac{1}{8}\right)^{\frac{1}{3}}$ , figure out what goes in the blank: $\left(? \right)^{3}=\dfrac{1}{8}$ To simplify $\left(\dfrac{1}{8}\right)^{\frac{1}{3}}$ , figure out what goes in the blank: $\left({\dfrac{1}{2}}\right)^{3}=\dfrac{1}{8}$ so $ \left(\dfrac{1}{8}\right)^{\frac{1}{3}}=\dfrac{1}{2}$ So $\left(\dfrac{1}{8}\right)^{\frac{4}{3}}=\left(\left(\dfrac{1}{8}\right)^{\frac{1}{3}}\right)^{4}=\left(\dfrac{1}{2}\right)^{4}$ $= \left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)\cdot \left(\dfrac{1}{2}\right)\cdot \left(\dfrac{1}{2}\right)$ $= \dfrac{1}{4}\cdot\left(\dfrac{1}{2}\right)\cdot \left(\dfrac{1}{2}\right)$ $= \dfrac{1}{8}\cdot\left(\dfrac{1}{2}\right)$ $= \dfrac{1}{16}$